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GEC 241- APPLIED II- WEEK ONE OF PARTICLES

INTRODUCTION TO MECHANICS

Mechanics is a branch of physical science that is concerned with the state of rest of of bodies subjected to the action of . mechanics is divided into two areas of study which are and dynamics.

From the first aspect of this study, we established that statics is concerned with the equilibrium of a body that is either at rest or moves at a constant . Here we will consider dynamics, which is the study that deals with the accelerated motion of a body.

The subject of dynamics will be presented in two parts:

i. Kinematics-which treats only the geometric aspects of the motion. ii. - which is the analysis of the forces causing the motion.

Dynamics is considered to be more involved than statics since both the forces applied to a body and its motion must be taken into account. There are many problems in engineering whose solutions require application of the principles of dynamics. Typically, the structural design of any vehicle such as automobile or airplane, requires consideration of motion to which it is subjected. This is also true for many mechanical devices, such as motors, pumps, movable tools, industrial manipulators and machinery. Moreover, predictions of the of artificial satellites, projectiles, and spacecraft are based on the theory of dynamics.

RECTILINEAR KINEMATICS: CONTINUOUS MOTION

Consider the kinematics of a particle that moves along a rectilinear or straight line path. Recall that a particle has a but negligible size and shape. Therefore we must limit application to those objects that have dimensions that are of no consequence in the analysis of the motion. In most problems, we will be interested in bodies of finite size such as rockets, projectiles or vehicles. Each of these objects can be considered as a particle, as long as the motion is characterized by the motion of its mass centre and any rotation of the body is neglected.

Rectilinear kinematics: The kinematics of a particle is characterized by specifying at any given instant, the particle’s position, velocity and .

Position: The origin O on the path is a fixed point and from this point the position coordinate s is used to specify the location of the particle at any given instant. The magnitude of s is the distance from O to the particle usually measured in meters (m) or feet (ft) and the sense of direction is defined by the algebraic sign on s. position is a vector quantity since it has both magnitude and direction. Fig. 1: Position

Displacement: The of the particle is defined as the change in the position. It can be either positive or negative depending on the initial position and the final position. The displacement of a particle is also a vector quantity which is different from distance. Specifically, distance travelled is a positive scalar that represents the total length of path over which the particle travels.

= ′ ∆𝑠𝑠 𝑠𝑠 − 𝑠𝑠

Fig. 2: Displacement

Velocity: It is the rate of change of displacement with taken. If the particle moves through a displacement during the time interval , the average velocity of the particle during this time interval is ∆𝑠𝑠 ∆𝑡𝑡 = ∆𝑠𝑠 𝑣𝑣𝑎𝑎𝑎𝑎 ∆𝑡𝑡 The instantaneous velocity is a vector defined as = lim 0( ), or

∆𝑡𝑡→ = 𝑣𝑣 (1) ∆𝑠𝑠⁄∆𝑡𝑡 𝒅𝒅𝒅𝒅 Its unit is m/s or ft/s. 𝒗𝒗 𝒅𝒅𝒅𝒅

Fig. 3: Velocity

Acceleration: The average acceleration of the particle during the time interval is defined as = ∆𝑣𝑣 𝑎𝑎𝑎𝑎𝑎𝑎 Where represents the difference in the ∆velocity𝑡𝑡 during the time interval , i.e. = ∆𝑣𝑣 ′ ∆𝑡𝑡 The instantaneous acceleration at time t is a vector∆𝑣𝑣 𝑣𝑣that− is𝑣𝑣 found by taking smaller and smaller values of

and corresponding smaller and smaller values of so that = lim 0( ) or

∆𝑡𝑡→ ∆ 𝑡𝑡 = ∆ 𝑣𝑣 (2) 𝑎𝑎 ∆𝑣𝑣⁄∆𝑡𝑡 𝒅𝒅𝒅𝒅 In terms of displacement and time, 𝒂𝒂the instantaneous𝒅𝒅𝒅𝒅 acceleration is

2 = 2 𝑑𝑑 𝑠𝑠 𝑎𝑎 𝑑𝑑𝑑𝑑

Fig. 4: Acceleration

Both the average and instantaneous acceleration can be either positive or negative. When the particle is slowing down, or its is decreasing, the particle is said to be decelerating. The unit is m/s2 or ft/s2.

Fig. 5: Deceleration

An important differential relation involving the displacement, velocity and acceleration along the path may be obtained by eliminating the time differential dt in eqns. (1) and (2), we have

= (3)

Constant Acceleration, =𝒂𝒂𝒂𝒂𝒂𝒂 𝒗𝒗𝒗𝒗𝒗𝒗

𝒄𝒄 When the acceleration 𝒂𝒂is constant,𝒂𝒂 each of the three kinematic equations can be integrated to obtain formulas that relate , , .

𝑐𝑐 𝑎𝑎 𝑣𝑣 𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡 Velocity as a Function of Time

Integrate = , assuming that initially , = 0 when = 0. 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑎𝑎 �𝑑𝑑𝑑𝑑 𝑣𝑣 𝑣𝑣 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 0 0 𝑐𝑐 �𝑣𝑣 𝑑𝑑𝑑𝑑 � 𝑎𝑎 𝑑𝑑𝑑𝑑 = + (4)

𝟎𝟎 𝒄𝒄 Position as a Function of Time 𝒗𝒗 𝒗𝒗 𝒂𝒂 𝒕𝒕

Integrate = = 0 + , assuming that initially = 0 when = 0 𝑑𝑑𝑑𝑑 𝑐𝑐 𝑣𝑣 �𝑑𝑑𝑑𝑑 𝑣𝑣 𝑎𝑎 𝑡𝑡 𝑠𝑠 𝑠𝑠 𝑡𝑡 = ( + ) 𝑠𝑠 𝑡𝑡 0 0 0 𝑐𝑐 �𝑠𝑠 𝑑𝑑𝑑𝑑 � 𝑣𝑣 𝑎𝑎 𝑡𝑡 𝑑𝑑𝑑𝑑 = + + (5) 𝟏𝟏 𝟐𝟐 𝟎𝟎 𝟎𝟎 𝒄𝒄 Velocity as a Function of Position 𝒔𝒔 𝒔𝒔 𝒗𝒗 𝒕𝒕 𝟐𝟐 𝒂𝒂 𝒕𝒕

It is either we solve for in eqn 4 and substitute into eqn 5 or integrate = , assuming that

= 0 = 0. 𝑡𝑡 𝑣𝑣𝑣𝑣𝑣𝑣 𝑎𝑎𝑐𝑐 𝑑𝑑𝑑𝑑 𝑣𝑣 𝑣𝑣 𝑎𝑎𝑎𝑎 𝑠𝑠 𝑠𝑠 = 𝑣𝑣 𝑠𝑠 0 0 𝑐𝑐 �𝑣𝑣 𝑣𝑣𝑣𝑣𝑣𝑣 �𝑠𝑠 𝑎𝑎 𝑑𝑑𝑑𝑑 = + ( ) (6) 𝟐𝟐 𝟐𝟐 𝟎𝟎 𝒄𝒄 𝟎𝟎 A typical example of constant𝒗𝒗 accelerated𝒗𝒗 𝟐𝟐𝒂𝒂 𝒔𝒔motion− 𝒔𝒔 occurs when a body falls freely towards the earth. If air resistance is neglected and the distance of fall is short, then the downward acceleration of the body when it is close to the earth is constant and approximately 9.81m/s2.

WORKED EXAMPLES:

1. The car In The figure below moves in a straight line such that for a short time its velocity is defined by = (3 2 + 2 ) / , where t is in seconds. Determine its position and acceleration when t=3 s. When = 0, s= 0. 𝑣𝑣 𝑡𝑡 𝑡𝑡 𝑚𝑚 𝑠𝑠 𝑡𝑡

Solution The position coordinate extends from the fixed origin O to the car, positive to the right. The car’s position can be determined from = , since this equation relates , . noting 𝒅𝒅𝒅𝒅 that s= 0 when t= 0 𝒗𝒗 𝒅𝒅𝒅𝒅 𝑣𝑣 𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡 = = ( + ) 𝒅𝒅𝒅𝒅 𝟐𝟐 𝒗𝒗 𝟑𝟑𝒕𝒕 𝟐𝟐𝟐𝟐 𝒅𝒅𝒅𝒅 = (3 2 + 2 ) 𝑠𝑠 𝑡𝑡 0 0

�𝑠𝑠 𝑑𝑑𝑑𝑑 � 𝑡𝑡 𝑡𝑡 𝑑𝑑𝑑𝑑 = 3 + 2

When t= 3 s, 𝑠𝑠 𝑡𝑡 𝑡𝑡

= 33 + 32 = 36

The acceleration is determined from = 𝑠𝑠 𝑚𝑚 𝒅𝒅𝒅𝒅 𝒂𝒂 𝒅𝒅𝒅𝒅 = = (3 2 + 2 ) 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑎𝑎 𝑡𝑡 𝑡𝑡 = 6t𝑑𝑑 +𝑑𝑑 2 𝑑𝑑𝑑𝑑

When t = 3 s,

A = 6(3) +2 = 20 m/s2

2. A small projectile is fired vertically downward into a fluid medium with an initial velocity of 60m/s. Due to the drag resistance of the fluid the projectile experiences a deceleration of = ( 0.4 3) / 2, where is in m/s. determine the projectile’s velocity and position 4 s after it is fired. 𝑎𝑎 SOLUTION − 𝑣𝑣 𝑚𝑚 𝑠𝑠 𝑣𝑣 Since the motion is downward, the position coordinate is positive downward To calculate the velocity, we will apply

= = 0.4 3 𝑑𝑑𝑑𝑑 𝑎𝑎 − 𝑣𝑣 Integrating with 0 = 60 / when = 0 𝑑𝑑𝑑𝑑

𝑣𝑣 𝑚𝑚 𝑠𝑠 𝑡𝑡 = 𝑣𝑣 0.4 3 𝑡𝑡 60 / 𝑑𝑑𝑑𝑑 0 � 𝑚𝑚 𝑠𝑠 � 𝑑𝑑𝑑𝑑 1 1− 1𝑣𝑣 2 𝑣𝑣 = 0 0.4 2 60 � � � 𝑡𝑡 − − − 𝑣𝑣

1 1 1 = 0.8 2 602 𝑡𝑡 � − � 𝑣𝑣 1 1 2 = + 0.8 / 602 − ⁄ 𝑣𝑣 � 𝑡𝑡� 𝑚𝑚 𝑠𝑠 When t = 4 s, = 0.559 /

For the position,𝑣𝑣 we will 𝑚𝑚use𝑠𝑠

1 1 2 = = + 0.8 602 − ⁄ 𝑑𝑑𝑑𝑑 𝑣𝑣 � 𝑡𝑡� 𝑑𝑑𝑑𝑑 1 1 2 = + 0.8 𝑠𝑠 𝑡𝑡 2 − ⁄ 0 0 60 � 𝑑𝑑𝑑𝑑 � � 𝑡𝑡� 𝑑𝑑𝑑𝑑 2 1 1 2 = + 0.8 𝑡𝑡 0.8 602 ⁄ 0 𝑠𝑠 � 𝑡𝑡� � 1 1 1 2 1 = + 0.8 0.4 602 ⁄ 60 𝑠𝑠 �� 𝑡𝑡� − � 𝑚𝑚 When t = 4 s, s = 4.43 m

3. During a test a rocket travels upwards at 75 m/s, and when it is 40 m from the ground its engine fails. Determine the maximum height reached by the rocket and its speed just before it hits the ground. While in motion, the rocket is subjected to a constant downward acceleration of 9.81 m/s2 due 𝑠𝑠𝐵𝐵 to . Neglect the effect of air resistance. SOLUTION For the maximum height, since the rocket is travelling upward, = +75 / when t= 0. At the 2 maximum height s = sB the velocity = 0. The acceleration = 9.81 / 𝑣𝑣𝐴𝐴 𝑚𝑚 𝑠𝑠 2 = 2 + 2 ( ) 𝑣𝑣𝐵𝐵 𝑎𝑎𝑐𝑐 − 𝑚𝑚 𝑠𝑠 0 = (75)2 + 2( 9.81)( 40 ) 𝑣𝑣𝐵𝐵 𝑣𝑣𝐴𝐴 𝑎𝑎𝑐𝑐 𝑠𝑠𝐵𝐵 − 𝑠𝑠𝐴𝐴

− 𝑠𝑠𝐵𝐵 − 𝑚𝑚 sB= 327 m To obtain the velocity of the rocket just before it hits the ground 2 = 2 + 2 ( ) = 0 + 2( 9.81)(0 327) 𝑣𝑣𝐶𝐶 𝑣𝑣𝐵𝐵 𝑎𝑎𝑐𝑐 𝑠𝑠𝐶𝐶 − 𝑠𝑠𝐵𝐵 = 80.1 / − −

𝑣𝑣𝐶𝐶 − 𝑚𝑚 𝑠𝑠 It may be defined as the angle described by a particle from one point to another, with respect to the time. Angular displacement is a vector quantity since it has both magnitude and direction.

C

B δϴ

ϴ O A Fig. 5: Angular displacement For example, let a line OB has its inclination ϴradians to the fixed line OA, as shown in the diagram above. If this line moves from OB to OC, through an angle δϴ during a short interval of time δt, then δϴ is known as the angular displacement of the line OB.

ANGULAR VELOCITY It may be defined as the rate of change of angular displacement with respect to time. It is usually expressed mathematically as = (7) 𝑑𝑑𝑑𝑑 Since it has magnitude and direction, therefore it is a vector quantity. Note that if the direction of the 𝜔𝜔 𝑑𝑑𝑑𝑑 angular displacement is constant, then the rate of change of magnitude of the angular displacement with respect with time is termed angular speed.

ANGULAR ACCELERATION It may be defined as the rate of change of with respect to time. Mathematically, is given as 2 = = = 2 (8) 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑 𝜃𝜃 It is also a vector quantity. 𝛼𝛼 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 �𝑑𝑑𝑑𝑑 � 𝑑𝑑𝑑𝑑

EQUATIONS OF ANGULAR MOTION It is important to note that the equations of angular motion are similar to that of the .

i. = 0 + . (9) 1 ii. = + . 2 (10) 𝜔𝜔 𝜔𝜔0 2𝛼𝛼 𝑡𝑡 2 = ( )2 + 2 . iii. 𝜃𝜃 𝜔𝜔 0 𝛼𝛼 𝑡𝑡 (11) ( + ) iv. = 0 (12) 𝜔𝜔 𝜔𝜔2 𝛼𝛼 𝜃𝜃 𝜔𝜔 𝜔𝜔 𝑡𝑡 𝜃𝜃 Where 0= Initial angular velocity in rad/s

𝜔𝜔 = Final angular velocity in rad/s

t𝜔𝜔 = Time in seconds

=Angular displacement in time t seconds

𝜃𝜃 = Angular acceleration in rad/s2

𝛼𝛼 Note: If a body is rotating at the rate of N r.p.m. (revolution per minute), then its angular velocity, 2 = 60 rad/s (14) 𝜋𝜋𝜋𝜋 RELATION BETWEEN LINEAR AND ANGULAR𝜔𝜔 QUANTITIES OF MOTION B Consider a body moving along a circular path from A to B Let r = radius of the circular path s

= angular displacement in radians

s = linear displacement O A 𝜃𝜃 𝜃𝜃 r v = linear velocity 𝜔𝜔 = angular velocity =linear acceleration 𝜔𝜔 = angular acceleration 𝑎𝑎 Fig. 6: Motion of a body along a circular path From the geometry, we know that 𝛼𝛼 = . (15) We also know that the linear velocity, ( . ) 𝑠𝑠 𝑟𝑟 𝜃𝜃 = = = × = . (16) 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑟𝑟 𝜃𝜃 𝑑𝑑𝑑𝑑 And linear acceleration, 𝑣𝑣 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑟𝑟 𝜔𝜔 ( . ) = = = × = . (17) 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑟𝑟 𝜔𝜔 𝑑𝑑𝑑𝑑 Example 𝑎𝑎 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑟𝑟 𝛼𝛼 1. A wheel accelerates uniformly from rest to 200 r.p.m in 20 seconds. What is its angular acceleration? How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.? Solution 2 ×2000 = = 209.5 Given: N0=0 0r = 0, N = 2000 r.p.m or 60 rad/s, t = 20 s 𝜋𝜋 Angular acceleration𝜔𝜔 can be calculated using𝜔𝜔 the formula

= 0 + .

𝜔𝜔 209𝜔𝜔.5 =𝛼𝛼0𝑡𝑡+ × 20 209.5 = = 10.475rad/s2 20 𝛼𝛼

Number of revolution𝛼𝛼 made by the wheel

( 0 + ) (0 + 209.5)20 = = = 2095 2 2 𝜔𝜔 𝜔𝜔 𝑡𝑡 Since the angular distance moved𝜃𝜃 by the wheel during one revolution is 𝑟𝑟2𝑟𝑟𝑟𝑟 radians, therefore the number of revolutions made by the wheel, 2095 𝜋𝜋 = = = 333.4 2 2 𝜃𝜃

𝑛𝑛 𝜋𝜋 𝜋𝜋

PRACTICE QUESTIONS 1. Initially, a car travels along a straight road with a speed of 35 m/s. if the brakes are applied and the speed of the car is reduced to 10 m/s in 15 s, determine the constant deceleration of the car. 2. The position of a particle is given by = (2 2 8 + 6) , where t is in seconds. Determine the time when the velocity of the particle is zero, and the total distance travelled by the particle 𝑠𝑠 𝑡𝑡 − 𝑡𝑡 𝑚𝑚 when t = 3 s. 3. A particle travels along a straight line with a velocity of = (20 0.05 2) m/s, where s is in meters. Determine the acceleration of the particle at s = 15 m. 𝑣𝑣 − 𝑠𝑠 4. A train starts from rest at a station and travels with a constant acceleration of 1m/s2. Determine the velocity of the train when t = 30s and the distance travelled during this time. 5. The position of a particle along a straight line is given by = (1.5 3 13.5 2 + 22.5 ) , where t is in seconds. Determine the position of the particle when t = 6 s and the total distance it 𝑠𝑠 𝑡𝑡 − 𝑡𝑡 𝑡𝑡 𝑓𝑓𝑓𝑓 travels during the 6s time interval. 6. The angular displacement of a body is a function of time and is given by equation: = 10 + 3 + 6 2, Determine the angular velocity, displacement and acceleration when t = 5 seconds. 𝜃𝜃 𝑡𝑡 𝑡𝑡 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠